A) \[\frac{{{\pi }^{2}}}{4}\]
B) \[{{\pi }^{2}}\]
C) \[\frac{5{{\pi }^{2}}}{4}\]
D) \[\frac{3{{\pi }^{2}}}{2}\]
Correct Answer: C
Solution :
As\[\vec{r}=(\vec{a}\times \vec{b})sinx+(\vec{b}\times \vec{c})cosy+2(\vec{c}\times \vec{a})\] \[\And \vec{r}.(\vec{a}+\vec{b}+\vec{c})=0\] \[\Rightarrow \]\[[\vec{a}\vec{b},\vec{c}](sinx+cosy+2)=0\] As\[[\vec{a}\vec{b},\vec{c}]\ne 0\therefore sin\,x+cos\,y=-2\] \[\therefore \]\[\sin x=-1\And \cos y=-1\] \[x=-\frac{\pi }{2}\And y=\pi \] \[{{x}^{2}}+{{y}^{2}}=\frac{5{{\pi }^{2}}}{4}\]
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